Answer:
Because the disturbances are in opposite directions for this superposition, the resulting amplitude is zero for pure destructive interference
Explanation:
If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.
20 J/s = 10% of Supply
20 J/s = (0.1) x (Supply)
Divide each side by 0.1:
Supply = (20 J/s) / (0.1)
<em>Supply = 200 J/s </em>(200 watts)
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Here's something to think about: What could you do to make the lamp more efficient ? Answer: Use it for a heater !
If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about. Suddenly ... bada-boom ... the lamp is 90% efficient !
Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Answer:
<em>The magnitude of the magnetic field will act in a direction towards me.</em>
<em></em>
Explanation:
When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. <em>The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field.</em> In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.
Answer:
50 N
4.2 N
Explanation:
i) The force needed to balance the boom is 2400 N. If the weight of the counterbalance is 2350 N, then the downward force the park attendant must apply is 50 N.
ii) When the boom is resting on the end support, the normal force is:
∑τ = Iα
-W (0.50) + F (3.0) − N (6.0) = 0
-0.50 W + 3.0 F = 6.0 N
N = (-0.50 W + 3.0 F) / 6.0
N = (-0.50 × 2350 + 3.0 × 400) / 6.0
N ≈ 4.2
